Contour integration - Branch cut I'm asked to show the following equality given $a\in (-1,1)\subset\Bbb R$
$$\int\limits_0^\infty\frac{x^a\ \log(x)}{(1+x)^2}dx=\frac{\pi\sin(\pi a)-a\pi^2\cos(\pi a)}{\sin^2(a\pi)}$$
So I'm trying to use the keyhole contour (as shown here) and so far I've been able to see the integrals along the circunferences tend to $0$ as $R\to\infty$ and $\varepsilon\to 0$. Computing $\text{Res}[f(z);-1]$ (where $f(z)$ equals the integrand) I get
$$\lim_{z\to -1}\frac{d}{dz}(1+z)^2f(z)=\lim_{z\to -1}az^{a-1}\log(z)+z^{a-1}=e^{\pi i(a-1)}(a(1+\pi i)+1)$$
On the other side,
$$\begin{align*}\int_\varepsilon^Rf(z)dz+\int_R^\varepsilon f(z)dz&=\int_\varepsilon^R\frac{z^a\log|z|}{(1+z)^2}dz+\int_R^\varepsilon\frac{z^a(\log|z|+2\pi i)}{(1+z)^2}dz\\ &=\int_\varepsilon^R\frac{z^a\log|z|}{(1+z)^2}dz+\int_R^\varepsilon\frac{z^a\log|z|}{(1+z)^2}dz-2\pi i\int_\varepsilon^R\frac{z^a}{(1+z)^2}\end{align*}$$
I know the result of the last integral, but I'm not sure whether what I've done is right, and what to do to finish it. Any help is appreciated.
 A: The point of the keyhole contour is to exploit the multivaluedness of the function being integrated.  In this case, the function is $z^a \log{z}$.  The right-hand side should look like
$$\int_0^{\infty} dx \frac{x^a \log{x}}{(1+x)^2} - e^{i 2 \pi a} \int_0^{\infty} dx \frac{x^a (\log{x}+i 2 \pi)}{(1+x)^2}$$
So you were halfway there - you had the $i 2 \pi$ from the log, but you also needed the other factor from the $z^a$.
The integral over the circular arcs vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.  It appears you are attempting to show this.  I will leave the rest of the details to the reader.
Now, I leave it to the reader to show that
$$\int_0^{\infty} dx \frac{x^a}{(1+x)^2} = \frac{\pi a}{\sin{\pi a}}$$
(which result you claim to have), so that
$$\left (1-e^{i 2 \pi a}\right) \int_0^{\infty} dx \frac{x^a \log{x}}{(1+x)^2} - i 2 \pi e^{i 2 \pi a} \frac{\pi a}{\sin{\pi a}} = i 2 \pi (-1) e^{i \pi a} (1+i \pi a)$$ 
(I get $1+i \pi a$, not $1+(1+i \pi)a$, in the residue.)
Do out the algebra - the sought result follows.
